Question

Find the solution of \[\;8\cos x\cos 2x\cos 4x = \dfrac{{\sin 6x}}{{\sin x}}\]

Answer 1

Hint: We will first cross multiply and rewrite the expression. Then, we will make the formula, $2\sin \theta \cos \theta = \sin 2\theta $ till we get a single term. Then, we will get the form, $\sin \alpha = \sin \beta $ and then apply the formula, $\alpha = n\pi + {\left( { - 1} \right)^n}\beta $ to find the solution of the equation.

Complete step-by-step answer:
We have to find the solution of \[\;8\cos x\cos 2x\cos 4x = \dfrac{{\sin 6x}}{{\sin x}}\]
Since, $\sin x$ is in the denominator, it cannot be equal to 0.
We will first cross multiply and rewrite the expression.
\[\;8\sin x\cos x\cos 2x\cos 4x = \sin 6x\]
We know that $2\sin \theta \cos \theta = \sin 2\theta $
Solving the LHS, we can write it as,
$
  \;4\left( {2\sin x\cos x} \right)\cos 2x\cos 4x = \sin 6x \\
   \Rightarrow 4\sin 2x\cos 2x\cos 4x = \sin 6x \\
$
Again we can form the formula of $\sin 2\theta $ in the LHS
$
  \;2\left( {2\sin 2x\cos 2x} \right)\cos 4x = \sin 6x \\
   \Rightarrow 2\sin 4x\cos 4x = \sin 6x \\
$
We can again form the formula,
$
  \;2\sin 4x\cos 4x = \sin 6x \\
   \Rightarrow \sin 8x = \sin 6x \\
$
Now, we know that if $\sin \alpha = \sin \beta \Rightarrow \alpha = n\pi + {\left( { - 1} \right)^n}\beta $
Then,
$8x = n\pi + {\left( { - 1} \right)^n}6x$
If $n$ is an even number, let $n = 2m$, where $m$ is an integer.
$
  8x = 2m\pi + 6x \\
   \Rightarrow 2x = 2m\pi \\
   \Rightarrow x = m\pi \\
$
Now, let $n$ is an odd number, let $n = 2m + 1$, where $m$ is an integer.
$
  8x = \left( {2m + 1} \right)\pi - 6x \\
   \Rightarrow 14x = \left( {2m + 1} \right)\pi \\
   \Rightarrow x = \dfrac{{\left( {2m + 1} \right)\pi }}{{14}} \\
$

Note: One should know the formulas of trigonometry to simplify the given expression. Also, one must have knowledge about the general solution when two trigonometric expressions are given equal. Here, students can also apply the formula of $2\cos A\cos B$ to simplify the expression.